6H2O + P4O10 = 4H3PO4
Coefficients: 6, 1, 4
Mean: the average. you have to add the values of the numbers and then divide by the amount of numbers there are. a common mistake to avoid is forgetting to divide the numbers at the end or subtracting them instead of adding.
mean: the middle number. you would first need to order the numbers from least to greatest. a common mistake to avoid is finding the middle number before ordering it from least to greatest
these two can also be commonly mistaken for one another because of the similar spelling.
Answer:
Number of valence electrons
Explanation:
Leftover: approximately 11.73 g of sulfuric acid.
<h3>Explanation</h3>
Which reactant is <em>in excess</em>?
The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a result,
- Al(OH)₃ is the limiting reactant.
- H₂SO₄ is in excess.
How many <em>moles</em> of H₂SO₄ is consumed?
Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
How many <em>grams</em> of H₂SO₄ is consumed?
The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.
How many <em>grams</em> of H₂SO₄ is in excess?
40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.
Answer:
8.33 hours
Explanation:
In order to solve this problem, we must apply Graham's law of diffusion in gases. Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its vapour density. For two gases we can write;
R1/R2=√d2/d1
Where;
R1= rate of diffusion of hydrogen
R2= rate diffusion of unknown gas
d1= vapour density of hydrogen
d2= vapour density of the unknown gas
Volume of hydrogen gas = 360cm^3
Time taken for hydrogen gas to diffuse= 1 hour =3600 secs
R1 = 360 cm^3/3600 secs = 0.1 cm^3 s-1
Vapour density of unknown gas = 25
Vapour density of hydrogen = 1
Substituting values,
0.1/R2 = √25/1
0.1/R2 = 5/1
5R2 = 0.1 × 1
R2 = 0.1/5
R2= 0.02 cm^3s-1
Volume of unknown gas = 600cm^3
Time taken for unknown gas to diffuse= volume of unknown gas/ rate of diffusion of unknown gas
Time taken for unknown gas to diffuse= 600/0.02
Time= 30,000 seconds or 8.33 hours
